Niuke.com High-Frequency Algorithm Question Series-BM4-Merge Two Sorted Linked Lists

Niuke.com High-Frequency Algorithm Question Series-BM4-Merge Two Sorted Linked Lists

Topic description

Input two increasing linked lists, the length of a single linked list is n, merge the two linked lists and make the nodes in the new linked list still in ascending order.

• Data range: 0 <= n <= 1000, -1000 <= node value <= 1000
• Requirements: space complexity O(1), time complexity O(n)

See the original title: BM4 merge two sorted linked lists

Solution 1: Linked List Traversal

• First, judge the special case, if the linked list 1 is empty, directly return the linked list 2; if the linked list 2 is empty, directly return the linked list 1.

• Otherwise, first declare a new fake head node, and then traverse linked list one and linked list two, the process is as follows:

  • If linked list one is empty, directly connect the remaining nodes of linked list two to the back, and terminate the traversal
  • If linked list 2 is empty, directly connect the remaining nodes of linked list 1 to the back, and terminate the traversal
  • Otherwise, compare the size of the current node of linked list 1 and linked list 2 to judge the next node, and then process the next node
• Finally, return the merged linked list.

code

 public class Bm004 {
    /**
     * 合并两个排序的链表
     *
     * @param list1 链表一
     * @param list2 链表二
     * @return
     */
    public static ListNode merge(ListNode list1, ListNode list2) {
        // 如果链表一为空，直接返回链表二
        if (list1 == null) {
            return list2;
        }
        // 如果链表二为空，直接返回链表一
        if (list2 == null) {
            return list1;
        }
        // 声明一个新的假头结点
        ListNode newList = new ListNode(-1), next = newList;
        // 遍历链表一和链表二
        while (list1 != null || list2 != null) {
            if (list1 == null) {
                // 如果链表一为空，直接将链表二剩下的结点接到后面，并终止遍历
                next.next = list2;
                break;
            } else if (list2 == null) {
                // 如果链表二为空，直接将链表一剩下的结点接到后面，并终止遍历
                next.next = list1;
                break;
            } else if (list1.val < list2.val) {  // 否则，比较链表一和链表二当前结点的大小，来判断下一个结点，然后处理下一个结点
                next.next = list1;
                list1 = list1.next;
            } else {
                next.next = list2;
                list2 = list2.next;
            }
            next = next.next;
        }
        // 最后，返回合并后的链表
        return newList.next;
    }

    public static void main(String[] args) {
        // 1 -> 3 -> 5
        ListNode list1 = ListNode.testCase3();
        System.out.println("链表一");
        ListNode.print(list1);
        // 2 -> 4 -> 6
        ListNode list2 = ListNode.testCase4();
        System.out.println("链表二");
        ListNode.print(list2);

        ListNode newList = merge(list1, list2);
        System.out.println("合并后的链表");
        ListNode.print(newList);
    }
}

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